Respuesta :
Answer:
62.89% probability that there will be smeared names on at most 4 t-shirts
Step-by-step explanation:
For each shirt, there are only two possible outcomes. Either there will be smeared names, or there will not. The probability of there being smeared names on a shirt is independent of other shirts. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
4% of the shirts are smeared:
So [tex]p = 0.04[/tex]
What is the probability, if a sample of 100 t-shirts is checked, there will be smeared names on at most 4 t-shirts?
This is [tex]P(X \leq 4)[/tex] when [tex]n = 100[/tex]. So
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.04)^{0}.(0.96)^{100} = 0.0169[/tex]
[tex]P(X = 1) = C_{100,1}.(0.04)^{1}.(0.96)^{99} = 0.0703[/tex]
[tex]P(X = 2) = C_{100,2}.(0.04)^{2}.(0.96)^{98} = 0.1450[/tex]
[tex]P(X = 3) = C_{100,3}.(0.04)^{3}.(0.96)^{97} = 0.1973[/tex]
[tex]P(X = 4) = C_{100,4}.(0.04)^{4}.(0.96)^{96} = 0.1994[/tex]
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0169 + 0.0703 + 0.1450 + 0.1973 + 0.1994 = 0.6289[/tex]
62.89% probability that there will be smeared names on at most 4 t-shirts
